Find the domain and range of each of the relations given below:

Question:

Find the domain and range of each of the relations given below:

(i) $R=\{(-1,1),(1,1),(-2,4),(2,4),(2,4),(3,9)\}$

(ii) $\mathrm{R}=\left\{\left(\mathrm{x}, \frac{1}{\mathrm{x}}\right): \mathrm{x}\right.$ is an interger, $\left.0<\mathrm{x}<5\right\}$

(iii) $R=\{(x, y): x+2 y=8$ and $x, y \in N\}$

(iv) $R=\{(x, y),: y=|x-1|, x \in Z$ and $|x| \leq 3\}$

 

Solution:

(i) Given: $R=\{(-1,1),(1,1),(-2,4),(2,4),(2,4),(3,9)\}$

$\operatorname{Dom}(R)=\{x:(x, y) \in R\}=\{-2,-1,1,2,3\}$

Range $(R)=\{y:(x, y) \in R\}=\{1,4,9\}$

(ii) Given:

$R=\left\{\left(x, \frac{1}{x}\right): x\right.$ is an interger, $\left.0

That means,

$R=\left\{(1,1),\left(2, \frac{1}{2}\right),\left(3, \frac{1}{3}\right),\left(4, \frac{1}{4}\right)\right\}$

$\operatorname{Dom}(R)=\left\{x:(x, y)^{\in} R\right\}=\{1,2,3,4\}$

Range $(R)=\left\{y:(x, y) \in_{R}\right\}=\left\{1, \frac{1}{2}, \frac{1}{3}, \frac{1}{4}\right\}$

(iii) Given: $\mathrm{R}=\{(\mathrm{x}, \mathrm{y}): \mathrm{x}+2 \mathrm{y}=8$ and $\mathrm{x}, \mathrm{y} \in \mathrm{N}\}$

That means, $R=\{(2,3),(4,2),(6,1)\}$

$\operatorname{Dom}(R)=\{x:(x, y) \in R\}=\{2,4,6\}$

Range $(R)=\left\{y:(x, y)^{\in} R\right\}=\{1,2,3\}$

(iv) Given: $R=\{(x, y): y=|x-1|, x \in Z$ and $|x| \leq 3\}$

$\operatorname{Dom}(R)=\left\{x:(x, y)^{\in} R\right\}=\{-3,-2,-1,0,1,2,3\}$

Range $(R)=\left\{y:(x, y)^{\in} R\right\}=\{0,1,2,3,4\}$

 

Leave a comment

Close

Click here to get exam-ready with eSaral

For making your preparation journey smoother of JEE, NEET and Class 8 to 10, grab our app now.

Download Now